http://www.algebrahelp.com/lessons/simplifying/foilmethod/pg2.htm . This link explain well on how to use the FOIL method.

anonymous

5 years ago

Okay

GoldPhenoix

5 years ago

Or, if you don't feel like reading the link, you can look at this pic. http://upload.wikimedia.org/wikipedia/commons/0/0c/MonkeyFaceFOILRule.JPG

GoldPhenoix

5 years ago

My bad. I mest up.

anonymous

5 years ago

okay thanks

GoldPhenoix

5 years ago

3(2x-1)(x+5) = 0
You want to solve X. So you want to distribute first. Do you know how to use the distributive property for: 3(2x-1) ?

anonymous

5 years ago

yea I'll try it

Jhannybean

5 years ago

Still need help solving this?...

reemii

5 years ago

I think that when the equation is in the form \(3(2x-1)(x+5) = 0\), it is better not to distribute. If a product of numbers is equal to 0, that means one of the numbers is zero.
Here, that means \(2x-1=0\) or \(x+5\). the solutions of both equations give the solutions of the initial equation.

anonymous

5 years ago

yes i need help still @Jhannybean @reemii

reemii

5 years ago

have yo found some solutions already?

reemii

5 years ago

what I said before is , to solve an equation of the form
\[f(x)g(h)h(x)=0\] means to "find the \(x\)'s".
A product of numbers equals zero if ONE OF the factors is zero.
tehrefore, solve separately the equations \(f(x)=0\), \(g(x)=0\), \(h(x)=0\) (etc).
Collect all the solutions you find for those equations, and this gives the solutions of the first equation.
In your case,
\(3(2x-1)(x+5)=0\)
You solve separately:
\(2x-1=0\) and \(x+5=0\).
For the first equation, the solutions is x = 1/2.
For the second equation, the solution is x = -5.
Collect them all: the solutions of your equation are 1/2 and -5.

reemii

5 years ago

In general, when facing an equation, people try to factorize, not to distribute. Factorizing allows to break the initial equation in smaller ones. Just as we (I) did.

reemii

5 years ago

@breja

Jhannybean

5 years ago

\[\large 3(2x-1)(x+5)=0\]\[\large 3(2x-1)=0\]divide 3 on both sides of the equation \[\large \frac{\cancel3(2x-1)}{\cancel 3}=\frac{0}{3}\]\[\large 2x-1 = 0\]\[\large 2x=1\]divide both sides by 2, you'll have 1 answer. \[\large x+3=0\]subtract -3 from both sides.\[\large x+3-3=0-3\] simplify nd you'll have another answer for x.